Freaks City

Problem A: Maximum Value Given two positive integers as input, output the larger value. Solution Simply compare the two values and output the maximum. #include <bits/stdc++.h> using namespace std; int main() { int a, b; cin >> a >> b; cout << max(a, b) << endl; return 0; } Problem B: Remove Vowel ...

Problem A: Textbook Availability Decision #include<iostream> using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(0); int regular, extra, discount; cin >> regular >> extra >> discount; double direct_cost = regular + extra * 0.5; double discounted_cost = (regular ...

Swapping Adjacent Nodes in Linked List This algorithm swaps every two adjacent nodes in a linked list using a dummy head approach for consistent handling. Key implementation details: Use a dummy head node to simplify edge cases Maintain current pointer before the pair being swapped Careful manage temporary references during swapping Ensure loo ...

What is a Segment Tree? A segment tree is a binary tree-based advanced data structure that supports flexible range operations. Unlike a Fenwick Tree (Binary Indexed Tree) which is limited to simple point update/range query use cases, segment trees can handle range updates with point queries, and even full range updates with range queries effici ...

Problem Statement Given n items, each with weight weight[i], value value[i], and group index group[i], pack them in to a knapsack with maximum capacity W. Each group can contain at most one item. Find the maximum total value. Solution 1: Two-Dimensional DP Array This is a classic grouped knapsack problem. First, we need to process all items sim ...

Legendary Players A direct mapping from player handles to their respective ratings is required. A hash map provides an efficient and clean way to resolve this without writing multiple conditional statements. #include <iostream> #include <string> #include <unordered_map> using namespace std; int main() { unordered_map< ...

Approach 1: Brute Force Method A straightforward solution involves checking all possible combinations of three indices using nested loops. This approach iterates through every possible triplet (i, j, k) in the array. The time compleixty is O(T × N³), which is impractical given the constraint 3 ≤ n ≤ 2 × 10⁵. This method would exceed time limits ...

24. Swap Nodes in Pairs Problem Link: LeetCode 24 Solution: This problem involves swapping adjacent nodes in a linked list in pairs. A dummy head node simplifies edge case handling, such as when the list has only one or two nodes. struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(nullptr) {} }; class Soluti ...

Problem B: Arbitrary Base Conversion This code converts a number from one arbitrary base to another. Implementation #include <stdio.h> #include <string.h> // Convert from base `src_base` string `src_num` to decimal integer. int convert_to_decimal(int src_base, const char *src_num) { int result = 0; int place_value = 1; ...

Unbounded Knapsack ProblemIn the classic 0/1 Knapsack problem, each item can be selected at most once. The Unbounded Knapsack problem modifies this constraint: each item can be chosen an unlimited number of times. Consider a knapsack with a maximum capacity of 4, and the following items:ItemWeightValueA115B320C430The core difference in implemen ...